12
$\begingroup$

Let $M_n(\mathbb{C})$ denote the algebra of $n \times n$ complex matrices, and $H_n(\mathbb{C})$ denote the linear space of $n \times n$ Hermitians. Both spaces are endowed with the usual operator norm. Assume that $$ \Phi \colon H_n(\mathbb{C}) \rightarrow H_n(\mathbb{C})$$ is a (real) linear map of norm $1.$

Then $\Phi$ has a natural linear extension to $M_n(\mathbb{C})$ by

$$\Phi(A) := \Phi\left({A+A^* \over 2}\right) + i\Phi\left({A-A^* \over 2i}\right).$$

Could you give me an example with $\|\Phi\| > 1 ?$ Or an explanation why this might happen?

$\endgroup$
5
  • $\begingroup$ The $\mathbb{C}$-linear extension of $\Phi$ as of above, let's denote it by $\tilde{\Phi}$. In the context of '$\mathbb{C}$omplexification' there's another look at it as a tensor product: $H_n(\mathbb{C})$ is an $\mathbb{R}$-linear subspace of $M_n(\mathbb{C})$ and every element in $M$ equals $h+ik$ with unique $h,k\in H$, hence $$M_n(\mathbb{C}) = \mathbb{C}\otimes_\mathbb{R}H_n(\mathbb{C}) \quad\text{and}\quad \tilde{\Phi} = \operatorname{id}\otimes_\mathbb{R}\Phi $$ If now $\|\text{id}\otimes_\mathbb{R}\Phi\|=\|\text{id}\|\cdot\|\Phi\|$ we were done. $\endgroup$
    – Hanno
    Feb 5 '17 at 19:09
  • 1
    $\begingroup$ This question is unclear. Operator norms do not exist independently. They are always induced by some norm. What is the norm that induces the so-called "usual" operator norm in question? If the inducing norm is the maximum absolute row sum norm, examples of $\|\Phi\|>1$ clearly exist (just let $\Phi(A)$ be the plain transpose of $A$); if the inducing norm is the Frobenius norm (hence the induced norm is the maximum singular value of an $n^2\times n^2$ matrix), the answer is clearly no; if the inducing norm is the maximum singular value, the answer is rather inobvious. $\endgroup$
    – user1551
    Mar 7 '17 at 15:19
  • $\begingroup$ @user1551: Good point. While I'm sure zoli will clarify, I hadn't thought of the potential ambiguity perhaps in part because of the tag "operator algebras," where the norm on $M_n(\mathbb C)$ considered as an algebra is usually by default the $C^*$-algebra one, or equivalently the maximum singular value, or equivalently the norm induced from acting on $\mathbb C^n$ with Euclidean norm. $\endgroup$ Mar 7 '17 at 15:45
  • 2
    $\begingroup$ Yes, the norm is the maximum singular value, and we have the Euclidean norm on $\mathbb{C}^n.$ For me, this is the usual operator norm on the matrix algebra $M_n(\mathbb{C}).$ Please, notice that in the commutative case, if we consider the $L^\infty[0,1]$ algebra ($[0,1]$ is equipped with the Lebesgue measure), the linear extension of $\Phi$ from real functions to complex functions does not increase the norm of $\Phi$ - this is probably a well-known result and goes back to M. Riesz, and A. E. Taylor. And the point is whether the same phenomena appears in the non-commutative setting, or not. $\endgroup$
    – zoli
    Mar 7 '17 at 16:30
  • $\begingroup$ I asked a question extending this one: math.stackexchange.com/questions/2185942 $\endgroup$ Mar 14 '17 at 7:18
6
$\begingroup$

I think I have an example showing the norm can be greater than $1$. For Hermitian $H=\begin{bmatrix}a&b\\\overline{b}&d\end{bmatrix}\in H_2(\mathbb C)$, define $\Phi(H)=\begin{bmatrix}0&\dfrac{a+di}{\sqrt2}\\\dfrac{a-di}{\sqrt2}&0\end{bmatrix}.$ We have $\|\Phi(H)\|=\frac1{\sqrt2}\sqrt{a^2+d^2}\leq\max\{|a|,|d|\}\leq\|H\|,$ with equality holding in case $|a|=|d|$ and $b=0$.

Consider the extension applied to $A=\begin{bmatrix}1&0\\0&i\end{bmatrix}$. We have $\|A\|=1$ while $\tilde{\Phi}(A)=\begin{bmatrix}0&0\\\sqrt2&0\end{bmatrix},$ so $\|\tilde{\Phi}(A)\|=\sqrt2$.


Motivation: To have the norm of the extension increase, it makes sense to look for cases where the real and imaginary parts of $A$ are "nonoverlapping," so that $A+A^*$ and $A-A^*$ can be as large as possible without increasing the norm of $A$. Then if we can have $\Phi$ map each of those parts in such a way that they do "overlap," an increase in norm can happen. This was done by taking distinct diagonal entries and placing them in the same off-diagonal positions. The off-diagonal was needed to allow the diagonal entries from the Hermitian matrices to be sent to real and imaginary parts, keeping the norm of $\Phi$ from being too large on $H_2(\mathbb C)$.

$\endgroup$
8
  • $\begingroup$ That's a nice example. $\endgroup$
    – zoli
    Mar 9 '17 at 22:49
  • 1
    $\begingroup$ I don't know if $\sqrt2$ is worst possible. $\endgroup$ Mar 10 '17 at 3:03
  • $\begingroup$ Yep, you are right. My bad. $\endgroup$ Mar 13 '17 at 16:59
  • $\begingroup$ @Jonas: You mean $\sqrt{2}$ is the worst in your example, or in general? $\endgroup$
    – zoli
    Mar 13 '17 at 18:00
  • $\begingroup$ @zoli: It is worst in my example, because for general $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$, $\|\tilde{\Phi}(A)\|=\frac{1}{\sqrt2}\max\{|a+id|,|a-id|\}\leq \frac{1}{\sqrt2}(|a|+|d|)\leq \sqrt2\max\{|a|,|d|\}\leq \sqrt2 \|A\|$. So in this case $\|\tilde{\Phi}\|=\sqrt 2$, but can it be that $\|\tilde{\Phi}\|>\sqrt2$ for other examples? $\endgroup$ Mar 13 '17 at 19:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.